Value of x obtained from equation left| {,begin{array}{*{20}{c}}{x + α }&β &gamma g

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3 and 4 .Determinants and Matrices

medium

The value of $x$ obtained from the equation $\left| {\,\begin{array}{*{20}{c}}{x + \alpha }&\beta &\gamma \\\gamma &{x + \beta }&\alpha \\\alpha &\beta &{x + \gamma }\end{array}\,} \right| = 0$ will be

$0 $ and $ - (\alpha + \beta + \gamma )$

$0$ and $(\alpha + \beta + \gamma )$

$1$ and $(\alpha - \beta - \gamma )$

$0 $ and $({\alpha ^2} + {\beta ^2} + {\gamma ^2})$

Solution

(a) Equation given, $\left| {\,\begin{array}{*{20}{c}}{x + \alpha + \beta + \gamma }&\beta &\gamma \\{x + \alpha + \beta + \gamma }&{x + \beta }&\alpha \\{x + \alpha + \beta + \gamma }&\beta &{x + \gamma }\end{array}\,} \right| = 0$,

$[{C_1} \to {C_1} + ({C_2} + {C_3})]$

or $(x + \alpha + \beta + \gamma )\,\left| {\,\begin{array}{*{20}{c}}1&\beta &\gamma \\1&{x + \beta }&\alpha \\1&\beta &{x + \gamma }\end{array}\,\,} \right|\, = 0$

or $(x + \alpha + \beta + \gamma )\,\left| {\,\begin{array}{*{20}{c}}1&\beta &\gamma \\0&x&{\alpha – \gamma }\\0&0&x\end{array}\,} \right|\, = \,0$,

$\left[ \begin{array}{l}{R_2} \to {R_2} – {R_1}\\{R_3} \to {R_3} – {R_1}\end{array} \right]$

or $(x + \alpha + \beta + \gamma )[{x^2} – 0] = 0$

or ${x^2}(x + \alpha + \beta + \gamma ) = 0$

$\therefore $ $x = 0$ or $x = – (\alpha + \beta + \gamma )$.

Standard 12

Mathematics

By using properties of determinants, show that:

$\left|\begin{array}{ccc}a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1\end{array}\right|=1+a^{2}+b^{2}+c^{2}$

hard

Let $D_1 =$ $\left| {\,\begin{array}{{20}{c}}a&b&{a + b}\\c&d&{c + d}\\a&b&{a – b}\end{array}\,} \right|$ and $D_2 =$ $\left| {\,\begin{array}{{20}{c}}a&c&{a + c}\\b&d&{b + d}\\a&c&{a + b + c}\end{array}\,} \right|$ then the value of $\frac{{{D_1}}}{{{D_2}}}$ where $b \ne 0$ and $ad \ne bc$, is

normal

Using properties of determinants, prove that:

$\left|\begin{array}{ccc}3 a & -a+b & -a+c \\ -b+a & 3 b & -b+c \\ -c+a & -c+b & 3 c\end{array}\right|=3(a+b+c)(a b+b c+c a)$

hard

If $\Delta = \left| {\,\begin{array}{{20}{c}}a&b&c\\x&y&z\\p&q&r\end{array}\,} \right|$, then $\left| {\,\begin{array}{{20}{c}}{ka}&{kb}&{kc}\\{kx}&{ky}&{kz}\\{kp}&{kq}&{kr}\end{array}\,} \right|$=

easy

Let the numbers $2, b, c$ be in an $A.P$ and $A = \left[ {\begin{array}{*{20}{c}}
1&1&1 \\
2&b&c \\
4&{{b^2}}&{{c^2}}
\end{array}} \right]$. If $det(A) \in [2,16]$ then $c$ lies in the interval

hard

(JEE MAIN-2019)

The value of $x$ obtained from the equation $\left| {\,\begin{array}{*{20}{c}}{x + \alpha }&\beta &\gamma \\\gamma &{x + \beta }&\alpha \\\alpha &\beta &{x + \gamma }\end{array}\,} \right| = 0$ will be

Solution

Similar Questions

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By using properties of determinants, show that:

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$\left|\begin{array}{ccc}3 a & -a+b & -a+c \\ -b+a & 3 b & -b+c \\ -c+a & -c+b & 3 c\end{array}\right|=3(a+b+c)(a b+b c+c a)$

If $\Delta = \left| {\,\begin{array}{{20}{c}}a&b&c\\x&y&z\\p&q&r\end{array}\,} \right|$, then $\left| {\,\begin{array}{{20}{c}}{ka}&{kb}&{kc}\\{kx}&{ky}&{kz}\\{kp}&{kq}&{kr}\end{array}\,} \right|$=

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